the value of a.
The expressions x4+x3-ax+11 and ax3+2x2–3 leave
the same remainder when divided by (x-2). Find
Answers
The Factor Theorem and The Remainder Theorem
Suppose we wish to find the zeros of f(x) = x
3 + 4x
2 − 5x − 14. Setting f(x) = 0 results in the
polynomial equation x
3 + 4x
2 − 5x − 14 = 0. Despite all of the factoring techniques we learned1
in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing
calculator, we get
The graph suggests that the function has three zeros, one of which is x = 2. It’s easy to show
that f(2) = 0, but the other two zeros seem to be less friendly. Even though we could use the
‘Zero’ command to find decimal approximations for these, we seek a method to find the remaining
zeros exactly. Based on our experience, if x = 2 is a zero, it seems that there should be a factor
of (x − 2) lurking around in the factorization of f(x). In other words, we should expect that
x
3 + 4x
2 − 5x − 14 = (x − 2) q(x), where q(x) is some other polynomial. How could we find such
a q(x), if it even exists? The answer comes from our old friend, polynomial division. Dividing
x
3 + 4x
2 − 5x − 14 by x − 2 gives
x
2 + 6x + 7
x−2 x
3 + 4x
2 − 5x − 14
−
x
3 −2x
2
6x
2 − 5x
−
6x
2 −12x)
7x − 14
− (7x −14)
0
As you may recall, this means x
3 + 4x
2 − 5x − 14 = (x − 2)
x
2 + 6x + 7
, so to find the zeros of f,
we now solve (x − 2)
x
2 + 6x + 7
= 0. We get x − 2 = 0 (which gives us our known zero, x = 2)
as well as x
2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic Formula to
get x = −3 ±
√
2. The point of this section is to generalize the technique applied here. First up is
a friendly reminder of what we can expect when we divide polynomia